求x^2-4xy 5y^2-2y 2013的最小值

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已知|2-x|与|x-y+4|互为相反数,求代数式3(x-y)-5(x-y)-3(x-y)+(y-x)-4(x+y)+3

|2-x|+|x-y+4|=02-x=0,x-y+4=0x=2,2-y+4=0,y=63(x-y)-5(x-y)^2-3(x-y)+(y-x)^2+4(x+y)^2+3(y-x)=-4(x-y)^2+

已知2x-y分之x+y=2,求代数式4x-2y分之x+y-4x+4y分之2x-y的值

2x-y分之x+y=2,则2x-y=2(x+y)4x-2y分之x+y-4x+4y分之2x-y=2(2x-y)分之x+y-4(x+y)分之2x-y=4(x+y)分之x+y-4(x+y)分之2(x+y)=

已知2x-y除以x+y等于2,求代数式(4x-2y除以x+y)-(4x+4y除以2x-y)的值

2X-Y/X+Y=2X+Y/2X-Y=1/2原式=[2(2X-Y)/X+Y]-[4(X+Y)/2X-Y]=2*2-4*(1/2)=4-2=2P.S.第一次为团队答题,好鸡东哦o(∩_∩)o有什么不懂的

2x-y=2,求[(x²+y²)-(x-y)²+2y(x-y)]÷4y

先化简[(x²+y²)-(x-y)²+2y(x-y)]/4y[(x²+y²)-(x-y)²+2y(x-y)]/4y=[x²+y&s

已知(2x-y)/(x+y)=2,求代数式(4x-2y)/(x+y)-(4x+4y)/(2x-y)的值.

(2x-y)/(x+y)=2两边同时乘以22*(2x-y)/(x+y)=2*2把2乘以进去,有(4x-2y)/(x+y)=4(2x-y)/(x+y)=2两边取倒数,有(x+y)/(2x-y)=1/2两

已知2x-y=10,求[(x²+y²)-(x-y)²+2y(x-y)]/4y

先化简[(x²+y²)-(x-y)²+2y(x-y)]/4y[(x²+y²)-(x-y)²+2y(x-y)]/4y=[x²+y&s

已知2x-y/x+3y=2,求代数式4x-2y/x+3y减4x+12y除以2x-y

2x-y/x+3y=2得x+3y/2x-y=1/24x-2y/x+3y减4x+12y除以2x-y=2(2x-y)/x+3y-4(x+3y)/2x-y=2乘以2-4乖以1/2=4-2=2

已知x/y=2,求2x(x+y)-y(x+y)/4x²-4xy+y²

原式=(2x-y)(x+y)/(2x-y)^2=(x+y)/(2x-y)x/y=2x=2y原式=3y/3y=1

若(x*x+y*y)(x*x+y*y)-4x*x*y*y=0,求代数式(x*x+5xy+y*y)/(x*x+2xy+y*

(x*x+y*y)(x*x+y*y)-4x*x*y*y=(x^4-2x^2y^2+y^4)=(x^2-y^2)^2=0x^2=y^2x/y=±1(x*x+5xy+y*y)/(x*x+2xy+y*y)=

已知2x-y/x+3y=2,求代数式4x-2y/x+3y减4x+12y/2x-y的值

答案是2.4x-2y/x+3y=(2x-y/x+3y)乘以2=4,4x+12y/2x-y=(x+3y/2x-y)乘以4=1/2x4=2,所以代数式等于4-2=2.

已知3x^2+xy-2y^2=0,求{(x+y)/(x-y)+4xy/(y^2-x^2)}/{(x+3y)*(x-y)}

3x^2+xy-2y^2=0推出(3x-2y)(x+y)=0推出x=-y或x=(2/3)y{(x+y)/(x-y)+4xy/(y^2-x^2)}/{(x+3y)*(x-y)}/x^2-9y^2推出:{

已知x*x-4xy+4y*y=0 求[2x(x+y)-y(x+y)]/(4x*x-4xy+y*y)的值?

即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²

已知x^2+4y^2-4x+4y+5=0,求(y^4-x^4)/(y-2x)(x+y)*2x-y

已知x^2+4y^2-4x+4y+5=0求((y^4-x^4)/(y-2x)(x+y))*((2x-y)/(xy-y^2))/((x^2+y^2)/y)的值答案:x²+4y²-4x

已知4x=9y求(1)x+y/y (2)y-x/2x

4x=9yx=9/4*y(1)(x+y)/y=[(9/4)y+y]/y=(9/4+1)y/y=9/4+1=13/4(2)(y-x)/2x=[y-(9/4)y]/[2*(9/4)y]=(1-9/4)y/

已知:X-Y\X+Y=4,求X-Y\2(X+Y)-4X+4Y\X-Y的值

(x-y)/(x+y)=4所以(x+y)/(x-y)=1/4(x-y)/[2(x+y)]-(4x+4y)/(x-y)=(1/2)*[(x-y)/(x+y)]-4[(x+y)/(x-y)]=(1/2)*

已知(2x-y)/(x+y)=2,求代数式(4x-2y)/(x+y)-(4x+4y)/(2x-y)的值

已知(2x-y)/(x+y)=2,则(4x-2y)/(x+y)=4,(x+y)/(2x-y)=1/2,(4x+4y)/(2x-y)=2(4x-2y)/(x+y)-(4x+4y)/(2x-y)=4-2=

y*y*y*y+2x*x*x*x=4x*x*y 求x等于几,y等于几,写出所有选项!

把原方程整理得:2x^4-4x^2y+y^4+1=0则这个关于x^2的一元二次方程有实数解,故得:(-4y)^2-8(y^4+1)≥0即有:(y^2-1)^2≤0当且仅当y^2-1=0时,上述方程有实

已知x²+y²+5=2x+4y,求【2x²-(x-y)(x-y)】【(x+y-1)(x-y

1,-3再问:过程。。。再答:★(x²-2x)+(y²-4y)=5★(x-1)²+(y-2)²=1+4-5★(x-l)²=0,(y-2)²=

若|x+2y-1|+y²+4y+4=0,求(2x-y)²-2(2x-y)(x+2y)+(x+2y)&

∵|x+2y-1|+y²+4y+4=0∴|x+2y-1|+(y+2)²=0∴x=5,y=-2(2x-y)²-2(2x-y)(x+2y)+(x+2y)²=[(2x

已知x=1/3,y=-1/2,求代数式x-(x+y)+(x+2y)-(x+3y)+(x+4y)-(x+5y)+...-(

原式=x-x+x-x+……-x+(2-1+4-3+5-4+……+2008-2007-2009)y=0+(1×1004-2009)y=-1005y=1005/2